2 ms·
Why do we need the property that X is compact? Say we define X (the square) to exclude the boundary: X = (0,1)x(0,1). This is no longer a compact set, but it se
by ridiculous_fish 2y ago
Why do we need the property that X is compact? Say we define X (the square) to exclude the boundary: X = (0,1)x(0,1). This is no longer a compact set, but it seems silly for the proof to now fail.
Is boundedness sufficient?
- NotYourLawyer 2y ago“Farthest” point doesn’t work if you drop compactness.
- lzralbu 2y agoNote that $A_k = \left[\frac{1}{k}, \frac{2}{k}\right] \times \left[\frac{1}{k}, \frac{2}{k}\right] \to \{(0, 0)\}$ in $[0,1]\times [0, 1]$ with the Hausdoff metric, so it is a Cauchy sequence in $X = (0,1) \times (0, 1)$ but without convergent subsequence, i.e., $F(X)$ is not complete, hence not compact so the finite subcover argument fails.